How do you write the partial fraction decomposition of the rational expression $(x^4+6)/(x^5+7x^3)$ ?

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Answer 1 · 2016-11-14T07:35:55

The answer is $=(-6/49)/x+(6/7)/x^3+(55/49x)/(x^2+7)$

$(x^4+6)/(x^5+7x^3)=(x^4+6)/((x^3)(x^2+7))$

Let's do the decompositiom in partial fractions

$(x^4+6)/((x^3)(x^2+7))=A/x+B/x^2+C/x^3+(Dx+E)/(x^2+7)$

$=(Ax^2(x^2+7)+Bx(x^2+7)+C(x^2+7)+x^3(Dx+E))/((x^3)(x^2+7))$

Therefore,

$x^4+6=Ax^2(x^2+7)+Bx(x^2+7)+C(x^2+7)+x^3(Dx+E)$

Let $x=0$ $=>$ $6=7C$ $=>$ $C=6/7$

Coefficients of $x^2$ , $=>$ $0=7A+C$ ; $A=-6/49$

coefficients of $x^4$ ; $=>$ ; $1=A+D$ ; $D=1+6/49=55/49$

Coefficients of $x^3$ ; $=>$ ; $0=B+E$

Coefficients of $x$ ; $=>$ ; $0=7B$ $=>$ $B=0$

$E=0$

$(x^4+6)/((x^3)(x^2+7))=(-6/49)/x+0/x^2+(6/7)/x^3+(55/49x+0)/(x^2+7)$

How do you write the partial fraction decomposition of the rational expression (x^4+6)/(x^5+7x^3) (x^4+6)/(x^5+7x^3)?